Let $h(x)=\dfrac{x-2}{\sqrt{x+7}-3}$ when $x\neq 2$. $h$ is continuous for all $x>-7$. Find $h(2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $6$ (Choice C) C $-3$ (Choice D) D $2$
Explanation: $\dfrac{x-2}{\sqrt{x+7}-3}$ is continuous for all $x>-7$ other than $x=2$, which means $h$ is continuous for all $x>-7$ other than $x=2$. In order for $h$ to also be continuous at $x=2$, the following equality must hold: $\lim_{x\to 2}h(x)=h(2)$ We will obtain the above equality by letting $h(2)=\lim_{x\to 2}h(x)$. So let's find $\lim_{x\to 2}h(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 2}h(x) \\\\ &=\lim_{x\to 2}\dfrac{x-2}{\sqrt{x+7}-3} \gray{\text{This is the rule for }x\neq 2} \\\\ &=\lim_{x\to 2}\dfrac{x-2}{\sqrt{x+7}-3}\cdot\dfrac{\sqrt{x+7}+3}{\sqrt{x+7}+3} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to 2}\dfrac{(x-2)(\sqrt{x+7}+3)}{x+7-3^2} \gray{\text{Simplify}} \\\\ &=\lim_{x\to 2}\dfrac{\cancel{(x-2)}(\sqrt{x+7}+3)}{\cancel{x-2}} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to 2}(\sqrt{x+7}+3) \\\\ &\text{(This is allowed because }x\neq 2) \\\\ &=\sqrt{2+7}+3 \gray{\text{Direct substitution}} \\\\ &=6 \end{aligned}$ We obtained that if we set $h(2)=6$, then $\lim_{x\to 2}h(x)=h(2)$, which makes $h$ continuous at $x=2$. Since we already saw that $h$ is continuous for any other $x>-7$, we can determine that it's continuous for all $x>-7$. In conclusion, $k=6$.